Answer:

Explanation:

First, we let #u=1+e^(2x)#. To integrate with respect to #u#, we divide by the derivative of #u#, which is #2e^(2x)#:#int\ e^x/(1+e^(2x))\ dx=1/2int\ e^x/(e^(2x)*u)\ du=1/2int\ e^x/(e^x*e^x*u)\ du=#

#=1/2int\ 1/(e^x*u)\ du#
To integrate with respect to #u#, we need everything expressed in terms of #u#, so we need to solve for what #e^x# is in terms of #u#:#u=1+e^(2x)#

#e^(2x)=u-1#

#2x=ln(u-1)#

#x=1/2ln(u-1)#

#x=ln((u-1)^(1/2))=ln(sqrt(u-1))#

#e^x=e^(ln(sqrt(u-1)))=sqrt(u-1)#

Now we can plug this back into the integral:#=1/2int\ 1/(e^x*u)\ du=1/2int\ 1/(sqrt(u-1)*u)\ du#

Next we will introduce a substitution with #z=sqrt(u-1)#. The derivative is:#(dz)/(du)=1/(2sqrt(u-1)#
so we divide by it to integrate with respect to #z# (remember that dividing is the same as multiplying by the reciprocal):#1/2int\ 1/(sqrt(u-1)*u)\ du=1/2int\ 1/(sqrt(u-1)*u)*2sqrt(u-1)\ dz=#

#=2/2int\ 1/u\ dz#

Now, we once again we have the wrong variable, so we need to solve for what #u# is equal to in terms of #z#:#z=sqrt(u-1)#

#u-1=z^2#

#u=z^2+1#

This gives:#int\ 1/u\ dz=int\ 1/(1+z^2)\ dz#

This is the common derivative of #tan^-1(z)#, so we get:#int\ 1/(1+z^2)\ dz=tan^-1(z)+C#

Undoing all the substitutions, we get:#tan^-1(z)+C=tan^-1(sqrt(u-1))+C=#